Question

ABC is right triangle right angled at C. If P is the length of the perpendicular from C to AB and a, b, c have the usual meaning, then \[\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}\] is……………..
A. \[\dfrac{1}{{{p}^{2}}}\]
B. \[\dfrac{2}{{{p}^{2}}}\]
C. \[{{p}^{2}}\]
D. \[2{{p}^{2}}\]

Answer 1

Hint: If two angles of a triangle have equal to the measures of two angles of another triangle, then the triangles are similar. In similar triangles the ratio of corresponding sides is equal.

Complete step by step solution:

Now, In triangle ABC and BCD
\[\angle \text{B = }\angle \text{B}\] (common angle in each triangle)
\[\angle \text{ACB = }\angle \text{CDB}\] (90 each angle)
By the property of similar triangles, two angles are common.
\[\therefore \Delta \text{ABC }\sim \text{ }\!\!\Delta\!\!\text{ BCD}\] (by Angle – Angle rule)
So, By basic proportionality ratio
\[\dfrac{AB}{BC}=\dfrac{AC}{CD}\]
\[\dfrac{c}{a}=\dfrac{b}{p}\]
\[c=\dfrac{ab}{p}\]
By Pythagoras theorem on\[\to A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}\]
\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\]
On putting the value of c
\[\dfrac{{{a}^{2}}{{b}^{2}}}{{{p}^{2}}}={{a}^{2}}+{{b}^{2}}\]
Dividing by \[{{a}^{2}}{{b}^{2}}\]to both sides, \[\dfrac{1}{{{p}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}\]

NOTE: Pythagoras attributed that the square of the hypotenuse of a right – angle triangle is equal to the sum of the squares on the other two sides.
When we prove two triangles are similar so we can only equate the ratio of corresponding sides of given triangles. So we need to choose corresponding sides carefully.